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Question Number 143932 by Ar Brandon last updated on 19/Jun/21

Answered by mathmax by abdo last updated on 19/Jun/21

$$1)\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{a}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}+{\int }_{\mathrm{a}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}\mathrm{for}\mathrm{alla}>0$$$$\mathrm{t}\to {\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{is}\mathrm{continue}\mathrm{on}[0,\mathrm{a}]\mathrm{so}\mathrm{integrable}\Rightarrow {\int }_0^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{t}2}\mathrm{dt}<\mathrm{\infty }$$$${\lim }_{\mathrm{t}\to +\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-{\mathrm{t}}^2}=0\Rightarrow {\int }_{\mathrm{a}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}<\mathrm{\infty }\Rightarrow \mathrm{I}\mathrm{is}\mathrm{cv}.$$$$$$

Answered by mathmax by abdo last updated on 19/Jun/21

$${\int }_{-1}^7\frac{\mathrm{dx}}{({}^3\sqrt{\mathrm{x}+1}})\mathrm{dx}{=}_{\mathrm{x}+1={\mathrm{t}}^3}{\int }_0^8\frac{{3\mathrm{t}}^2\mathrm{dt}}{\mathrm{t}}=3{\int }_0^8\mathrm{tdt}=3{\left[\frac{{\mathrm{t}}^2}{2}\right]}_0^8=\frac{3}{2}\times 8^2\Rightarrow$$$$\mathrm{ths}\mathrm{integral}\mathrm{is}\mathrm{cv}.$$

Answered by mathmax by abdo last updated on 19/Jun/21

$${\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\sqrt{\mathrm{t}}}}{\mathrm{t}}\mathrm{dt}{=}_{\sqrt{\mathrm{t}}=\mathrm{y}}{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{y}}}{{\mathrm{y}}^2}\left(2\mathrm{y}\right)\mathrm{dy}=2{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{y}}}{\mathrm{y}}\mathrm{dy}\mathrm{for}ϵ>0$$$${\lim }_{\mathrm{y}\to +\mathrm{\infty }}{\mathrm{y}}^{1+ϵ}\frac{{\mathrm{e}}^{-\mathrm{y}}}{\mathrm{y}}={\lim }_{\mathrm{y}\to +\mathrm{\infty }}{\mathrm{y}}^{ϵ}{\mathrm{e}}^{-\mathrm{y}}=0\Rightarrow \mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{cv}.$$

Answered by mathmax by abdo last updated on 19/Jun/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{logt}}{\mathrm{t}+2}\mathrm{dt}\Rightarrow \mathrm{I}{=}_{\mathrm{t}+2=\mathrm{x}}{\int }_2^{\mathrm{\infty }}\frac{\log (\mathrm{x}-2)}{\mathrm{x}}\mathrm{dx}\mathrm{for}\xi >0$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^{1+\xi }\frac{\log (\mathrm{x}-2)}{\mathrm{x}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^{\xi }\log (\mathrm{x}-2)=+\mathrm{\infty }$$$$\Rightarrow \mathrm{I}\mathrm{diverges}$$

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